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PROBLEM STATEMENT:

Find all functions $f :\mathbb{R}^{+}_{0} \to \mathbb{R}^{+}_{0} $ satisfying the functional equation $$f\bigl(f(x)-x\bigr)=2x.$$

MY PROGRESS:

This question seemed pretty tough to me. All I could figure out was:

  1. $f(x) = 2x$ and $f(x)=-x$ (considering the domain and co-domain to be all reals) satisfy the given equation.
  2. $f\bigl(f(0)\bigr)=0$.
  3. By substituting certain values for $x$, I found out that $$f\bigl(n f(0)\bigr)= (-n+1) f(0)$$ for $n=1,3,-5,11, \dots$.

But these didn't help me much. What should be my approach? Clearly the general way of substitution and doing some algebraic manipulations didn't help you to proceed with the question. Any hints or ideas?

Also the solution to this question used some sort of sequences and recurrence relation that I couldn't understand at all. Any ideas in this respect?

Thanks.

Feel free to add tags and a better title as I am not good at that.

PS:

In general $$f\bigl(f(x)-nx\bigr)=(n+1)x,$$ for all integers $n$, is satisfied by $f(x)=(n+1)x$. Can this be proved explicitly?

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    $\begingroup$ For the functional equation to make sense, we must have $f(x)-x \geq 0$. I would try playing around a bit with that. $\endgroup$ Commented Feb 13, 2022 at 13:43
  • $\begingroup$ @Marktmeister have you come up with something? $\endgroup$
    – AbVk1718
    Commented Feb 14, 2022 at 8:59
  • $\begingroup$ @AbVk1718 It's generally not a good idea to add a new question (or changing the original question) to the post, especially after getting an answer to the original one. The recommended practice is posting a new one, with giving a link to the previous question for reference. But in this case, as the more general question you added later was very much related to the original one, and the same method could be applied to it, I added some explanation at the end of my original post answering that. $\endgroup$ Commented Feb 14, 2022 at 20:37
  • $\begingroup$ @MohsenShahriari I will take care of that from next time $\endgroup$
    – AbVk1718
    Commented Feb 15, 2022 at 0:29

1 Answer 1

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As Marktmeister has mentioned in a comment, for the functional equation $$ f \bigl ( f ( x ) - x \bigr ) = 2 x \tag 0 \label 0 $$ to make sense for a given $ f : \mathbb R ^ + _ 0 \to \mathbb R ^ + _ 0 $ and all $ x \in \mathbb R ^ + _ 0 $, one must have $ f ( x ) \ge x $ for all $ x \in \mathbb R ^ + _ 0 $. In particular, $ f \bigl ( f ( x ) - x \bigr ) \ge f ( x ) - x $ for all $ x \in \mathbb R ^ + _ 0 $, which by \eqref{0} gives $ f ( x ) \le 3 x $ for all $ x \in \mathbb R ^ + _ 0 $. Let's iterate this method: define the sequences $ ( a _ n ) _ { n = 0 } ^ \infty $ and $ ( b _ n ) _ { n = 0 } ^ \infty $ of positive real numbers recursively by $ a _ 0 = 1 $ and for any nonnegative integer $ n $, $ b _ n = \frac 2 { a _ n } + 1 $ and $ a _ { n + 1 } = \frac 2 { b _ n } + 1 $. Note that similar to what we did above, by \eqref{0} and mathematical induction, one gets $ a _ n x \le f ( x ) \le b _ n x $ for all $ x \in \mathbb R ^ + _ 0 $ and all positive integers $ n $. Therefore, if we prove that as $ n $ tends to infinity, $ a _ n $ increasingly goes to $ 2 $ and $ b _ n $ decreasingly goes to $ 2 $, then we end up with $ f ( x ) = 2 x $ for all $ x \in \mathbb R ^ + _ 0 $, which indeed gives a solution to \eqref{0}. For this purpose, first use mathematical induction to prove that for any nonnegative integer $ n $, $ 1 \le a _ n < 2 $ and $ 2 < b _ n \le 3 $. Then, note that for any nonnegative integer $ n $, $$ 0 < 2 - a _ { n + 1 } = 2 - \left ( \frac 2 { \frac 2 { a _ n } + 1 } + 1 \right ) = \frac { 2 - a _ n } { 2 + a _ n } \le \frac { 2 - a _ n } { 2 + 1 } = \frac { 2 - a _ n } 3 $$ and $$ 0 < b _ { n + 1 } - 2 = \left ( \frac 2 { \frac 2 { b _ n } + 1 } + 1 \right ) - 2 = \frac { b _ n - 2 } { b _ n + 2 } < \frac { b _ n - 2 } { 2 + 2 } = \frac { b _ n - 2 } 4 \text . $$ Hence, by mathematical induction, we get $$ 0 < 2 - a _ n \le \frac { 2 - a _ 0 } { 3 ^ n } = \frac 1 { 3 ^ n } $$ and $$ 0 < b _ n - 2 \le \frac { b _ 0 - 2 } { 4 ^ n } = \frac 1 { 4 ^ n } $$ for all nonnegative integers $ n $, which proves what was desired.


Concerning the part you added later, if $ f : \mathbb R ^ + _ 0 \to \mathbb R ^ + _ 0 $ satisfies $$ f \bigl ( f ( x ) - \alpha x \bigr ) = \beta x \tag 1 \label 1 $$ for some constants $ \alpha , \beta \in \mathbb R ^ + $ and all $ x \in \mathbb R ^ + _ 0 $, the same method as above works. Note that for all $ x \in \mathbb R ^ + _ 0 $ and all $ \gamma \in \mathbb R ^ + $, $ f \bigl ( f ( x ) - \alpha x \bigr ) \ge \gamma \bigl ( f ( x ) - \alpha x \bigr ) $ implies $ f ( x ) \le \left ( \frac \beta \gamma + \alpha \right ) x $ (by \eqref{1}), and $ f \bigl ( f ( x ) - \alpha x \bigr ) \le \gamma \bigl ( f ( x ) - \alpha x \bigr ) $ implies $ f ( x ) \ge \left ( \frac \beta \gamma + \alpha \right ) x $. Therefore, similar to before, we can define $ ( a _ m ) _ { m = 0 } ^ \infty $ and $ ( b _ m ) _ { m = 0 } ^ \infty $ by $ a _ 0 = \alpha $, $ b _ m = \frac \beta { a _ m } + \alpha $ and $ a _ { m + 1 } = \frac \beta { b _ m } + \alpha $, and inductively show that $ a _ m x \le f ( x ) \le b _ m x $ for all $ x \in \mathbb R ^ + $ and all nonnegative integers $ m $. Taking limits, you can then find out that $ f ( x ) = c x $ gives the unique solution, where $ c $ is the positive root of $ c ^ 2 - \alpha c - \beta $, i.e. $ c = \frac { \alpha + \sqrt { \alpha ^ 2 + 4 \beta } } 2 $. I leave verifying the details to you. In the special case where $ \alpha = n $ and $ \beta = n + 1 $, we have $ c = n + 1 $.

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  • $\begingroup$ What is the motivation behind defining the sequences $a_n$ and $b_n$? $\endgroup$
    – AbVk1718
    Commented Feb 14, 2022 at 9:09
  • $\begingroup$ I am a bit confused with the induction part. Shouldn't you have assumed that $0 <2- a_{n}$ to prove that $0<2-a_{n+1}$? $\endgroup$
    – AbVk1718
    Commented Feb 14, 2022 at 9:31
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    $\begingroup$ @AbVk1718 The motivation: if we know that $ 2 x = f \bigl ( f ( x ) - x \bigr ) \le c \bigl ( f ( x ) - x \bigr ) $ for some constant $ c $, we get $ f ( x ) \ge \left ( \frac 2 c + 1 \right ) x $; the same happens when swapping $ \le $ and $ \ge $. For your other question, note that $ 0 < 2 - a _ n $ is already known for all $ n $, from the previous step ($ 1 \le a _ n < 2 $ and $ 2 < b _ n \le 3 $), which was proven by induction separately; the same is true for $ b _ n - 2 > 0 $. $\endgroup$ Commented Feb 14, 2022 at 12:10
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    $\begingroup$ Possible motivation: if you consider the function $g(x)=f(x)-x$, then $g(g(x))+g(x)=2x$. By iterating $x\to g(x)\to g^2(x)\to \cdots$, you can think of this new functional equation as being akin to a sort of recurrence relation. The fact it's over $\mathbb{R}^+$ lets you conclude $g(x)=x$. $\endgroup$
    – user574848
    Commented Feb 14, 2022 at 14:30
  • $\begingroup$ @user574848 I agree that the method you suggested is a natural way of thinking about this problem. But as it was indicated in the original post that the owner of the post had problems with a previous solution containing recurrences, I guessed that iterating the unknown function $ f $ had been the source of their problem, and tried presenting a solution avoiding that. There is still recursion involved in defining $ a _ n $ and $ b _ n $, but the recursive step is not given by applying $ f $ (or other unknowns like your $ g $) to the previous terms, but only by a simple algebraic expression. $\endgroup$ Commented Feb 14, 2022 at 20:47

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